Skellam distribution

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Skellam
Probability mass function
Examples of the probability mass function for the Skellam distribution.
Examples of the probability mass function for the Skellam distribution. The horizontal axis is the index k. (The function is only defined at integer values of k. The connecting lines do not indicate continuity.)
Parameters μ10,μ20
Support k{,2,1,0,1,2,}
pmf e(μ1+μ2)(μ1μ2)k/2Ik(2μ1μ2)
Mean μ1μ2
Median N/A
Variance μ1+μ2
Skewness μ1μ2(μ1+μ2)3/2
Kurtosis 1μ1+μ2
MGF e(μ1+μ2)+μ1et+μ2et
CF e(μ1+μ2)+μ1eit+μ2eit

The Skellam distribution is the discrete probability distribution of the difference N1N2 of two statistically independent random variables N1 and N2, each Poisson-distributed with respective expected values μ1 and μ2. It is useful in describing the statistics of the difference of two images with simple photon noise, as well as describing the point spread distribution in sports where all scored points are equal, such as baseball, hockey and soccer.

The distribution is also applicable to a special case of the difference of dependent Poisson random variables, but just the obvious case where the two variables have a common additive random contribution which is cancelled by the differencing: see Karlis & Ntzoufras (2003) for details and an application.

The probability mass function for the Skellam distribution for a difference K=N1N2 between two independent Poisson-distributed random variables with means μ1 and μ2 is given by:

p(k;μ1,μ2)=Pr{K=k}=e(μ1+μ2)(μ1μ2)k/2Ik(2μ1μ2)

where Ik(z) is the modified Bessel function of the first kind. Since k is an integer we have that Ik(z)=I|k|(z).

Derivation

The probability mass function of a Poisson-distributed random variable with mean μ is given by

p(k;μ)=μkk!eμ.

for k0 (and zero otherwise). The Skellam probability mass function for the difference of two independent counts K=N1N2 is the convolution of two Poisson distributions: (Skellam, 1946)

p(k;μ1,μ2)=n=p(k+n;μ1)p(n;μ2)=e(μ1+μ2)n=max(0,k)

Since the Poisson distribution is zero for negative values of the count (p(N<0;μ)=0), the second sum is only taken for those terms where n0 and n+k0. It can be shown that the above sum implies that

p(k;μ1,μ2)p(k;μ1,μ2)=(μ1μ2)k

so that:

p(k;μ1,μ2)=e(μ1+μ2)(μ1μ2)k/2I|k|(2μ1μ2)

where I k(z) is the modified Bessel function of the first kind. The special case for μ1=μ2(=μ) is given by Irwin (1937):

p(k;μ,μ)=e2μI|k|(2μ).

Using the limiting values of the modified Bessel function for small arguments, we can recover the Poisson distribution as a special case of the Skellam distribution for μ2=0.

Properties

As it is a discrete probability function, the Skellam probability mass function is normalized:

k=p(k;μ1,μ2)=1.

We know that the probability generating function (pgf) for a Poisson distribution is:

G(t;μ)=eμ(t1).

It follows that the pgf, G(t;μ1,μ2), for a Skellam probability mass function will be:

G(t;μ1,μ2)=k=p(k;μ1,μ2)tk=G(t;μ1)G(1/t;μ2)=e(μ1+μ2)+μ1t+μ2/t.

Notice that the form of the probability-generating function implies that the distribution of the sums or the differences of any number of independent Skellam-distributed variables are again Skellam-distributed. It is sometimes claimed that any linear combination of two Skellam distributed variables are again Skellam-distributed, but this is clearly not true since any multiplier other than ±1 would change the support of the distribution and alter the pattern of moments in a way that no Skellam distribution can satisfy.

The moment-generating function is given by:

M(t;μ1,μ2)=G(et;μ1,μ2)=k=0tkk!mk

which yields the raw moments mk . Define:

Δ =def μ1μ2
μ =def (μ1+μ2)/2.

Then the raw moments mk are

m1=Δ
m2=2μ+Δ2
m3=Δ(1+6μ+Δ2)

The central moments M k are

M2=2μ,
M3=Δ,
M4=2μ+12μ2.

The mean, variance, skewness, and kurtosis excess are respectively:

E(n)=Δ,σ2=2μ,γ1=Δ/(2μ)3/2,γ2=1/2.

The cumulant-generating function is given by:

K(t;μ1,μ2) =def ln(M(t;μ1,μ2))=k=0tkk!κk

which yields the cumulants:

κ2k=2μ
κ2k+1=Δ.

For the special case when μ1 = μ2, an asymptotic expansion of the modified Bessel function of the first kind yields for large μ:

p(k;μ,μ)14πμ[1+n=1(1)n{4k212}{4k232}{4k2(2n1)2}n!23n(2μ)n].

(Abramowitz & Stegun 1972, p. 377). Also, for this special case, when k is also large, and of order of the square root of 2μ, the distribution tends to a normal distribution:

p(k;μ,μ)ek2/4μ4πμ.

These special results can easily be extended to the more general case of different means.

Bounds on weight above zero

If XSkellam(μ1,μ2), with μ1<μ2, then

exp((μ1μ2)2)(μ1+μ2)2e(μ1+μ2)2μ1μ2e(μ1+μ2)4μ1μ2Pr{X0}exp((μ1μ2)2)

Details can be found in Poisson distribution

References

See also